Question: What is the extraneous solution to these equations? $\dfrac{x^2 - x}{x + 2} = \dfrac{2x + 4}{x + 2}$
Multiply both sides by $x + 2$ $ \dfrac{x^2 - x}{x + 2} (x + 2) = \dfrac{2x + 4}{x + 2} (x + 2)$ $ x^2 - x = 2x + 4$ Subtract $2x + 4$ from both sides: $ x^2 - x - (2x + 4) = 2x + 4 - (2x + 4)$ $ x^2 - x - 2x - 4 = 0$ $ x^2 - 3x - 4 = 0$ Factor the expression: $ (x - 4)(x + 1) = 0$ Therefore $x = 4$ or $x = -1$ The original expression is defined at $x = 4$ and $x = -1$, so there are no extraneous solutions.